1. **Problem statement:** Given the quadratic equation $\alpha x^2 + \beta x + \gamma = 0$ with $\alpha \neq 0$, prove that the sum of the roots $x_1 + x_2 = -\frac{\beta}{\alpha}$.\n\n2. **Formula and rules:** For a quadratic equation $ax^2 + bx + c = 0$, the roots $x_1$ and $x_2$ satisfy:\n$$x_1 + x_2 = -\frac{b}{a} \quad \text{and} \quad x_1 x_2 = \frac{c}{a}.$$\nThese come from the factorization of the quadratic as $a(x - x_1)(x - x_2) = 0$.\n\n3. **Proof:** Starting from the quadratic equation:\n$$\alpha x^2 + \beta x + \gamma = 0,$$\nwe can write it as:\n$$\alpha (x - x_1)(x - x_2) = 0.$$\nExpanding the right side:\n$$\alpha (x^2 - (x_1 + x_2)x + x_1 x_2) = 0,$$\nwhich gives:\n$$\alpha x^2 - \alpha (x_1 + x_2) x + \alpha x_1 x_2 = 0.$$\nComparing coefficients with the original equation, we get:\n$$\beta = -\alpha (x_1 + x_2) \quad \Rightarrow \quad x_1 + x_2 = -\frac{\beta}{\alpha}.$$\n\n4. **Explanation:** The sum of the roots is directly related to the coefficient of $x$ in the quadratic equation. This relationship holds for any quadratic with $\alpha \neq 0$.\n\n**Final answer:**\n$$x_1 + x_2 = -\frac{\beta}{\alpha}.$$